Complex Analysis 29 | Liouville's Theorem
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- Опубликовано: 7 фев 2025
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While studing C*-algebras, i found a pretty interesting proof of non-emptyness of the spectrum of a element without the Liouville theorem. It uses only a trick deriving under the integral curve and a Hahn-Banach extension, but the argument used can be used to derive the Liouville theorem as consequence!
At 7:30 it's true that if a holomorphic function f is s.t. f'(z) = 0 everywhere, then this implies that f is constant, but the proof is more subtle than in real analysis. In real analysis, we can just apply the Mean Value Theorem: f(x) - f(x_0) = f'(c)(x - x_0) = 0 which means f(x) = f(x_0) for all x. But in complex analysis, we need to use the fact that since f is holomorphic, we can express it locally as a power series in a disk around any point z_0, with coefficients involving f(z_0), f'(z_0), f''(z_0), and so on. Since f'(z) = 0 for all z, this means all higher derivatives also vanish, reducing the power series to f(z) = f(z_0). Therefore, f is constant in the whole disk. But this holds no matter how large the disk is (because no matter the disk, f is holomorphic in it with f'(z)=0)! Therefore we can conclude that f is constant over the entire plane C.
Actually a same proof as in real analysis also works here (without using the mean value theorem) since the definition of the derivative is the same. It's a similar thing we do in multivariable calculus.
@@brightsideofmaths What proof? I don't think you can use the Mean Value Theorem.
Also instead of the mean value theorem, you can apply the mean value inequality which also holds for complex-valued functions. You can also google for "A complex Rolle’s theorem"
@@individuoenigmatico1990 Yes, not the mean value theorem but the mean value inequality.
@@brightsideofmaths Yes, you are correct. But we have to remember that this inequality is only guaranteed in convex open sets. I never noticed it, but we can actually apply the contour integral inequality that you proved in Chapter 19 to prove the "mean value inequality" for complex-valued functions. I write this long comment hoping someone may find it useful in the future.
Let f be holomorphic in an open set U and let γ:[a,b]->U be a piecewise continuously differentiable curve in U, then we know, by the results of Chapter 28, that f' is itself continuous in U and therefore that ∫_γ f'(z) dz is well defined. Then, by the results of Chapter 19, we have that || ∫_γ f'(z) dz || ≤ max_(z∈γ) {||f'(z)||} • lenght(γ). But since f' has by definition f as antiderivative in U, then - by the results of Chapter 20 - we have that:
||f(γ(b))-f(γ(a))|| ≤ max_(z∈γ) {||f'(z)||} • lenght(γ)
If we now take U to be a convex open set then, given any two points z1,z2∈U, we know U contains the straight line γ:[0,1]->U such that γ(t)=z1+t(z2-z1)
We then have γ(1)=z2, γ(0)=z1, and lenght(γ)=||z2-z1||. Therefore, by applying the previous result, we obtain:
||f(z2)-f(z1)|| ≤ max_(z∈γ) {||f'(z)||} • ||z2-z1||.
This proves the "mean value inequality". So if f is an holomorphic function in a convex open set U we get ||f(z2)-f(z1)|| ≤ max_(z∈γ) {||f'(z)||} • ||z2-z1|| for all z1,z2∈U.
So if f'(z)=0 everywhere in the convex open set U, then ||f(z2)-f(z1)||=0 for all z1,z2∈U and therefore f is constant in U, which is the result we intented to prove. So you were correct in saying that we could use the "mean value inequality" to prove the result we wanted to prove.
This has, by the way, a nice and neat corollary: in convex open sets antiderivatives are determined up to a constant. In other words: if we let F(z) and G(z) be antiderivatives of f(z) in a convex open set U, then there exists a constant z0∈C such that F(z)=G(z)+z0 for all z€U. In fact (F-G)'(z)=f(z)-f(z)=0 everywhere in U, so for the previous result we know that F-G is a constant function in U. So there exists a constant z0∈C such that F(z)=G(z)+z0 for all z€U.
can you do a series on abstract algebra?
Yes, I can :)
Yay! Can’t wait!
Hello Julian! Great video as always. I have a question, though. At 7:00, you state that "because our domain of the function f(z) is so large that every disk lies inside it." Which seems to contradict the fact that we are talking of "entire function," so in this case, "the derivatives of a holomorphic function cannot be arbitrarily large." A little clarification would be helpful. Thank you in advance. Mario
Thanks for the question. For an entire function, every disc lies completely in the domain. So the formula holds for every r > 0.
@@brightsideofmaths Thank you for your fast answer. So that is why the circle must be inside the domain D?
@@mariolemelin27 That is the requirement of Cauchy's inequality, indeed :)
@@brightsideofmaths Great. Thank you 🙂
🎯 Key Takeaways for quick navigation:
00:00 📚 Introduction to Koshy's Inequalities and Liouville's Theorem
- Introduction to Koshy's Inequalities and Liouville's Theorem.
- Thanks to channel supporters for making these videos possible.
- Formulation of important inequalities assuming an open set D, holomorphic function f on D, and a closed disk brz0 inside D.
01:42 📊 Inequality Estimation for Derivatives
- Derivation and discussion of an estimate for the absolute value of the nth derivative of a holomorphic function.
- Use of parametrized curves, Contour integrals, and manipulation of the integral expression.
- Application of standard estimates for integrals, resulting in the bound of the absolute value.
04:22 🧐 Key Steps in the Inequality Proof
- Demonstration of crucial steps inthe inequality proof.
- Utilization of the triangle inequality for integrals.
- Ignoring a complex number's absolute value due to it being always one.
05:46 🎯 Application of the Inequality to Entire Functions
- Introduction and application of the inequality to entire functions.
- Definition of entire functions and the assumption of boundedness.
- Estimation of the supremum of the absolute value of an entire function along a circle.
07:22 📈 Liouville's Theorem and Its Implications
- Application of the estimate to the first derivative, leading to Liouville's Theorem.
- Implication that a holomorphic, bounded, and non-constant entire function implies a restricted domain.
- Recognition of the significance of Liouville's Theorem in identifying constant entire functions.
08:31 🤓 Closing Remarks and Preview
- Recapitulation of Liouville's Theorem and its application to identifying non-constant, bounded, holomorphic functions.
- Mention of revisiting Liouville's Theorem in future applications.
- Conclusion and farewell to the viewers.
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"Koshy's Inequalities" should mean "Cauchy's inequalities" ...
can you explain where in the previous lectures we did
"if derivative of holomorphic function on connected domain is zero then function is constant"
It follows quickly from the definition :)
Why did the answer of the third question in the quiz be no? Constant functions in C seem to meet all the conditions.
Thanks! This was a typo in the question :)